Problem: What is the extraneous solution to these equations? $\dfrac{x^2 - 2}{x + 4} = \dfrac{4x - 6}{x + 4}$
Explanation: Multiply both sides by $x + 4$ $ \dfrac{x^2 - 2}{x + 4} (x + 4) = \dfrac{4x - 6}{x + 4} (x + 4)$ $ x^2 - 2 = 4x - 6$ Subtract $4x - 6$ from both sides: $ x^2 - 2 - (4x - 6) = 4x - 6 - (4x - 6)$ $ x^2 - 2 - 4x + 6 = 0$ $ x^2 + 4 - 4x = 0$ Factor the expression: $ (x - 2)(x - 2) = 0$ Therefore $x = 2$ The original expression is defined at $x = 2$ and $x = 2$, so there are no extraneous solutions.